symfony:我们不能拥有隐藏的实体字段吗？

Question

我正在使用symfony中的实体字段渲染表单.

当我选择常规实体字段时,它运行良好.

$builder
    ->add('parent','entity',array(
            'class' => 'AppBundle:FoodAnalytics\Recipe',
            'attr' => array(
                'class' => 'hidden'
            )
        ))

当我选择 - > add('parent','hidden')时会抛出以下错误:

表单的视图数据应该是标量,数组或\ ArrayAccess的实例,但是类AppBundle\Entity\FoodAnalytics\Recipe的实例.您可以通过将"data_class"选项设置为"AppBundle\Entity\FoodAnalytics\Recipe"或添加视图转换器来将类AppBundle\Entity\FoodAnalytics\Recipe的实例转换为标量,数组或实例来避免此错误ArrayAccess接口.500内部服务器错误 - LogicException

我们不能有隐藏的实体字段吗？为什么不？我是否有义务将另一个隐藏字段用于检索实体ID？

编辑:

基本上,我正在尝试做的是在显示它之前水合表单,但阻止用户更改其中一个字段(这里的父级).这是因为我需要将Id作为参数传递,我不能在表单操作URL中执行此操作.

Answer 1

我认为你只是对字段类型以及它们各自代表的内容感到困惑.

一个entity字段是一个类型的choice字段.选择字段旨在包含用户在表单中可选择的值.呈现此表单时,Symfony将根据实体字段的基础类生成可能选择的列表,列表中每个选项的值是相应实体的ID.提交表单后,Symfony将为您呈现代表所选实体的对象.该entity字段通常用于呈现实体关联(例如,roles您可以选择分配给a 的列表user).

如果您只是尝试为实体的ID字段创建占位符,那么您将使用hidden输入.但这只适用于您正在创建的表单类表示实体(即表单是data_class指您已定义的实体).然后,ID字段将正确映射到由表单定义的类型的实体的ID data_class.

编辑:下面描述的特定情况的一种解决方案是创建一个新的字段类型(让我们称之为EntityHidden)扩展hidden字段类型,但处理数据转换以转换为/从实体/ id.这样,您的表单将包含实体ID作为隐藏字段,但是一旦提交表单,应用程序就可以访问实体本身.当然,转换由数据转换器执行.

以下是后代的这种实现示例:

namespace My\Bundle\Form\Extension\Type;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\Form\DataTransformerInterface;

/**
 * Entity hidden custom type class definition
 */
class EntityHiddenType extends AbstractType
{
    /**
     * @var DataTransformerInterface $transformer
     */
     private $transformer;

    /**
     * Constructor
     *
     * @param DataTransformerInterface $transformer
     */
    public function __construct(DataTransformerInterface $transformer)
    {
        $this->transformer = $transformer;
    }

    /**
     * @inheritDoc
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        // attach the specified model transformer for this entity list field
        // this will convert data between object and string formats
        $builder->addModelTransformer($this->transformer);
    }

    /**
     * @inheritDoc
     */
    public function getParent()
    {
        return 'hidden';
    }

    /**
     * @inheritDoc
     */
    public function getName()
    {
        return 'entityhidden';
    }
}

请注意,在表单类型类中,您所要做的就是将隐藏的实体分配给其对应的表单字段属性(在表单模型/数据类中),Symfony将使用实体的ID正确生成隐藏的输入HTML它的价值.希望有所帮助.

Answer 2

刚刚在Symfony 3上做了这个,并意识到它与已经发布的内容有点不同所以我认为值得分享.

我刚刚制作了一个通用数据转换器,可以在所有表​​单类型中轻松重用.你必须传递你的表单类型,就是这样.无需创建自定义表单类型.

首先让我们来看看数据转换器:

<?php

namespace AppBundle\Form;

use Doctrine\Common\Persistence\ObjectManager;
use Symfony\Component\Form\DataTransformerInterface;
use Symfony\Component\Form\Exception\TransformationFailedException;

/**
 * Class EntityHiddenTransformer
 *
 * @package AppBundle\Form
 * @author  Francesco Casula <fra.casula@gmail.com>
 */
class EntityHiddenTransformer implements DataTransformerInterface
{
    /**
     * @var ObjectManager
     */
    private $objectManager;

    /**
     * @var string
     */
    private $className;

    /**
     * @var string
     */
    private $primaryKey;

    /**
     * EntityHiddenType constructor.
     *
     * @param ObjectManager $objectManager
     * @param string        $className
     * @param string        $primaryKey
     */
    public function __construct(ObjectManager $objectManager, $className, $primaryKey)
    {
        $this->objectManager = $objectManager;
        $this->className = $className;
        $this->primaryKey = $primaryKey;
    }

    /**
     * @return ObjectManager
     */
    public function getObjectManager()
    {
        return $this->objectManager;
    }

    /**
     * @return string
     */
    public function getClassName()
    {
        return $this->className;
    }

    /**
     * @return string
     */
    public function getPrimaryKey()
    {
        return $this->primaryKey;
    }

    /**
     * Transforms an object (entity) to a string (number).
     *
     * @param  object|null $entity
     *
     * @return string
     */
    public function transform($entity)
    {
        if (null === $entity) {
            return '';
        }

        $method = 'get' . ucfirst($this->getPrimaryKey());

        // Probably worth throwing an exception if the method doesn't exist
        // Note: you can always use reflection to get the PK even though there's no public getter for it

        return $entity->$method();
    }

    /**
     * Transforms a string (number) to an object (entity).
     *
     * @param  string $identifier
     *
     * @return object|null
     * @throws TransformationFailedException if object (entity) is not found.
     */
    public function reverseTransform($identifier)
    {
        if (!$identifier) {
            return null;
        }

        $entity = $this->getObjectManager()
            ->getRepository($this->getClassName())
            ->find($identifier);

        if (null === $entity) {
            // causes a validation error
            // this message is not shown to the user
            // see the invalid_message option
            throw new TransformationFailedException(sprintf(
                'An entity with ID "%s" does not exist!',
                $identifier
            ));
        }

        return $entity;
    }
}

所以我们的想法是你通过在那里传递对象管理器,你想要使用的实体然后是字段名来获取实体ID来调用它.

基本上是这样的:

new EntityHiddenTransformer(
    $this->getObjectManager(),
    Article::class, // in your case this would be FoodAnalytics\Recipe::class
    'articleId' // I guess this for you would be recipeId?
)

我们现在就把它们放在一起吧.我们只需要表单类型和一些YAML配置,然后我们就可以了.

<?php

namespace AppBundle\Form;

use AppBundle\Entity\Article;
use Doctrine\Common\Persistence\ObjectManager;
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\Form\Extension\Core\Type\HiddenType;
use Symfony\Component\Form\Extension\Core\Type\SubmitType;
use Symfony\Component\OptionsResolver\OptionsResolver;

/**
 * Class JustAFormType
 *
 * @package AppBundle\CmsBundle\Form
 * @author  Francesco Casula <fra.casula@gmail.com>
 */
class JustAFormType extends AbstractType
{
    /**
     * @var ObjectManager
     */
    private $objectManager;

    /**
     * JustAFormType constructor.
     *
     * @param ObjectManager $objectManager
     */
    public function __construct(ObjectManager $objectManager)
    {
        $this->objectManager = $objectManager;
    }

    /**
     * @return ObjectManager
     */
    public function getObjectManager()
    {
        return $this->objectManager;
    }

    /**
     * {@inheritdoc}
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('article', HiddenType::class)
            ->add('save', SubmitType::class);

        $builder
            ->get('article')
            ->addModelTransformer(new EntityHiddenTransformer(
                $this->getObjectManager(),
                Article::class,
                'articleId'
            ));
    }

    /**
     * {@inheritdoc}
     */
    public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults([
            'data_class' => 'AppBundle\Entity\MyEntity',
        ]);
    }
}

然后在你的services.yml文件中:

app.form.type.article:
    class: AppBundle\Form\JustAFormType
    arguments: ["@doctrine.orm.entity_manager"]
    tags:
        - { name: form.type }

在你的控制器中:

$form = $this->createForm(JustAFormType::class, new MyEntity());
$form->handleRequest($request);

而已 :-)

Answer 3

在 Symfony 5 中，我使用实现 DataTransformerInterface 接口的隐藏类型的解决方案。

<?php

namespace App\Form\Type;

use Doctrine\Persistence\ManagerRegistry;
use Symfony\Component\Form\DataTransformerInterface;
use Symfony\Component\Form\Exception\TransformationFailedException;
use Symfony\Component\Form\Extension\Core\Type\HiddenType;
use Symfony\Component\Form\FormBuilderInterface;

/**
 * Defines the custom form field type used to add a hidden entity
 *
 * See https://symfony.com/doc/current/form/create_custom_field_type.html
 */
class EntityHiddenType extends HiddenType implements DataTransformerInterface
{

    /** @var ManagerRegistry $dm */
    private $dm;

    /** @var string $entityClass */
    private $entityClass;

    /**
     *
     * @param ManagerRegistry $doctrine
     */
    public function __construct(ManagerRegistry $doctrine)
    {
        $this->dm = $doctrine;
    }

    /**
     *
     * {@inheritdoc}
     */
    public function buildForm(FormBuilderInterface $builder, array $options): void
    {
        // Set class, eg: App\Entity\RuleSet
        $this->entityClass = sprintf('App\Entity\%s', ucfirst($builder->getName()));
        $builder->addModelTransformer($this);
    }

    public function transform($data): string
    {
        // Modified from comments to use instanceof so that base classes or interfaces can be specified
        if (null === $data || !$data instanceof $this->entityClass) {
            return '';
        }

        $res = $data->getId();

        return $res;
    }

    public function reverseTransform($data)
    {
        if (!$data) {
            return null;
        }

        $res = null;
        try {
            $rep = $this->dm->getRepository($this->entityClass);
            $res = $rep->findOneBy(array(
                "id" => $data
            ));
        }
        catch (\Exception $e) {
            throw new TransformationFailedException($e->getMessage());
        }

        if ($res === null) {
            throw new TransformationFailedException(sprintf('A %s with id "%s" does not exist!', $this->entityClass, $data));
        }

        return $res;
    }
}

并在表单中使用该字段：

use App\Form\Type\EntityHiddenType;

public function buildForm(FormBuilderInterface $builder, array $options): void
{
    // Field name must match entity class, eg 'ruleSet' for App\Entity\RuleSet
    $builder->add('ruleSet', EntityHiddenType::class);
}

Answer 4

这可以通过表单主题相当干净地实现，使用标准hidden字段主题代替实体的主题。我认为使用转换器可能是矫枉过正，因为隐藏和选择字段将提供相同的格式。

{% block _recipe_parent_widget %}
    {%- set type = 'hidden' -%}
    {{ block('form_widget_simple') }}
{% endblock %}